A uniform thin rod of mass 0.950 kg and length 1.20 m is mounted on a
horizontal pin through the center of the rod at right angles to the rod such that the rod is
free to spin in a vertical plane about the pin without friction. The rod is positioned
horizontally at rest by a person who then releases the rod. The rod remains at rest until a
ball of mass 0.265 kg is dropped on the rod from a height of above theℎ0 = 1. 400 𝑚
rod. The ball hits the rod at a point exactly between the center of the rod and one end of
the rod. The ball bounces straight up to a maximum height of 0.599 m above the point of
contact. How fast is the rod spinning after the ball hits the rod the first time but before the
ball hits the rod again? (Note that the moment of inertia of a uniform thin rod, with
respect to an axis that passes through the midpoint of the rod and is perpendicular to the
rod, is given by where M is the mass of the rod and L is the length of the rod.)
1
12 𝑀𝐿2


To solve this, we use the principles of **conservation of angular momentum** and **energy considerations**.

---

### Step 1: Information and Formulas
#### Given:
1. **Rod mass (\(M\))**: \(0.950 \, \text{kg}\)
2. **Rod length (\(L\))**: \(1.20 \, \text{m}\)
3. **Ball mass (\(m\))**: \(0.265 \, \text{kg}\)
4. **Height from which the ball is dropped (\(h_0\))**: \(1.400 \, \text{m}\)
5. **Height the ball reaches after collision (\(h_1\))**: \(0.599 \, \text{m}\)
6. **Moment of inertia of the rod about its center**: 
   \[
   I_\text{rod} = \frac{1}{12} M L^2
   \]

#### To Find:
Angular velocity (\(\omega\)) of the rod after the collision.

---

### Step 2: Linear Momentum of the Ball Before and After Collision
The **velocity of the ball before impact** is calculated using:
\[
v_\text{before} = \sqrt{2 g h_0}
\]
The **velocity of the ball after impact** is calculated using:
\[
v_\text{after} = \sqrt{2 g h_1}
\]
Where \(g = 9.8 \, \text{m/s}^2\).

---

### Step 3: Angular Momentum Conservation
The system's angular momentum about the pivot is conserved. Initially, only the ball has angular momentum since the rod is stationary.

#### Initial Angular Momentum (Before Collision):
\[
L_\text{initial} = m v_\text{before} \cdot \frac{L}{4}
\]
where \(\frac{L}{4}\) is the perpendicular distance from the pivot to the point of impact.

#### Final Angular Momentum (After Collision):
After the collision, the rod rotates with angular velocity \(\omega\), and the ball continues with linear velocity \(v_\text{after}\). The total angular momentum is:
\[
L_\text{final} = I_\text{rod} \cdot \omega + m v_\text{after} \cdot \frac{L}{4}
\]

---

### Step 4: Conservation Equation
Equating the initial and final angular momenta:
\[
m v_\text{before} \cdot \frac{L}{4} = I_\text{rod} \cdot \omega + m v_\text{after} \cdot \frac{L}{4}
\]
Substitute \(I_\text{rod} = \frac{1}{12} M L^2\) and solve for \(\omega\):
\[
\omega = \frac{m v_\text{before} \cdot \frac{L}{4} - m v_\text{after} \cdot \frac{L}{4}}{\frac{1}{12} M L^2}
\]

---

### Step 5: Substituting Values
1. \(v_\text{before} = \sqrt{2 \cdot 9.8 \cdot 1.400} = 5.242 \, \text{m/s}\)
2. \(v_\text{after} = \sqrt{2 \cdot 9.8 \cdot 0.599} = 3.430 \, \text{m/s}\)
3. \(I_\text{rod} = \frac{1}{12} \cdot 0.950 \cdot (1.20)^2 = 0.114 \, \text{kg·m}^2\)

Substitute into the equation:
\[
\omega = \frac{0.265 \cdot (5.242 - 3.430) \cdot \frac{1.20}{4}}{0.114}
\]

Simplify:
\[
\omega = \frac{0.265 \cdot 1.812 \cdot 0.30}{0.114} = \frac{0.144 \, \text{kg·m}^2/\text{s}}{0.114 \, \text{kg·m}^2} = 1.26 \, \text{rad/s}
\]

---

### Final Answer:
The angular velocity of the rod is:
\[
\boxed{1.26 \, \text{rad/s}}
\]

---

Let me know if you want any further clarification or additional steps. Here are some related questions to explore:

1. How does the height the ball reaches after collision affect the angular velocity of the rod?
2. What would happen if the ball hit the rod at a different point?
3. How would adding friction at the pivot change the results?
4. What if the rod had a different shape or moment of inertia?
5. Can energy conservation be used as an alternate method in this scenario?

**Tip:** Always check whether a problem involves angular or linear quantities to apply the correct conservation laws

A uniform thin rod of mass 0.950 kg and length 1.20 m is mounted on a horizontal pin through the center of the rod at right angles to the rod such that the rod is free to spin in a vertical plane about the pin without friction. The rod is positioned horizontally at rest by a person who then releases the rod. The rod remains at rest until a ball of mass 0.265 kg is dropped on the rod from a height of 1.400 m above the rod. The ball hits the rod at a point exactly between the center of the rod and one end of the rod. The ball bounces straight up to a maximum height of 0.599 m above the point of contact. How fast is the rod spinning after the ball hits the rod the first time but before the ball hits the rod again? (Note that the moment of inertia of a uniform thin rod, with respect to an axis that passes through the midpoint of the rod and is perpendicular to the rod, is given by (1/12)ML^2.)

Learn how to calculate the angular velocity of a uniform thin rod after a collision with a falling ball using conservation of angular momentum and energy principles. This physics problem involves rotational dynamics, moment of inertia, and linear motion. Suitable for undergraduate physics students.

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